Question

Anyone good with large derivatives and want to help out?

Answer 1

try me

Answer 2

ok, the derivative is Mo*V/C^2(1-(v^2/c^2))^3/2 where Mo=mass of object at rest, c=speed of light and v=speed

Answer 3

V IS CONSTANT ?

Answer 4

no v is the variable for speed, it's the only non constant

Answer 5

I wish I could represent this better, but I can't use the equation editor worth a darn, is there any other complications you see with my equations?

Answer 6

what value do they want you to use for speed of light, or are they having you solve the function in terms of speed of light as a constant C?

Answer 7

just solve with c, I need to find the derivative of the equation above...

Answer 8

so far I see you use the product rule for the numerator.....the product rule for the denominator, followed by the chain rule then the quotient rule....then quotient rule for the whole thing

Answer 9

that is exactly what you do

Answer 10

yup...Im having trouble with simple algebra of fixing the variables up at the end

Answer 11

I can post the last step that I am sure of

Answer 12

MV squared over C use product rule, thenv squared over c squared product rule... yeah do that

Answer 13

i keep tripping myself up on the variables, funny i can solve one of these big guys when its full of decimals and garbage but simple variables trip me up haha

Answer 14

ok here's what I got..... {[C^2(1-(v^2/c^2))^3/2]+[3MoV(1-(v^2/c^2))^1/2]}/ (C^2(1-(v^2/c^2))^3/2)^3

Answer 15

haha, if that isn't clear let me know...it's the [C^2...+ 3MoV...] divided by the C^2...

Answer 16

the variables trip me up too man

Answer 17

ok I'm just going to throw out small parts of it that I'm unsure of

Answer 18

is (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^3?

Answer 19

im pretty sure those are not the same, no

Answer 20

right sorry I meant (C^2(1-(v^2/c^2))^3/2)^3 the same as C^4*(1-(v^2/c^2)^2...last number should have been a 2

Answer 21

where are you getting C to the 4th up front from?

Answer 22

(C^2)^2....that last ^2 applies to both in the denominator doesnt it?

Answer 23

which last ^2 are you talking about? theres a ^3 at the end and the v squared over c squared is in parenthesis so im not sure what your saying (text based math formulas suck)

Answer 24

yeah it does, I'm going to give up on this one for now, thanks for the help tho

Answer 25

sorry i couldnt help more, ill give it a try again after i make some food

Answer 26

you deriave with respect to what v ?

Answer 27

yes, v is independent, m(v) is dependent

Answer 28

if my equation confuses you, it's already the derivative of Einstein's equation m(v) mass of a moving particle, now we have to derive it again